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First determine when a group automorphism can carry a transposition to another type of involution. An involution in $S_n$ is a product of $k$ disjoint transpositions, and its conjugacy class has sizeAn automorphism preserves element orders and conjugacy-class sizes. Equality with the transposition-class size $n(n-1)/2$ requiresFor $k=2$ this says $(n-2)(n-3)=4$, which has no integral solution $n\ge 4$. For $k=3$, the minimum value of the left side, at $n=6$, is $4!=24=2^2\cdot 3!$; it increases strictly for larger $n$. For $k\ge 4$, even its minimum $(2k-2)!$ exceeds $2^{k-1}k!$: at $k=4$ this is $720>192$, and the ratio of these two quantities increases at the next step by $(2k)(2k-1)/(2(k+1))>1$. Thus only in degree six can the transposition class be exchanged with a different involution class, namely the triple transpositions.

We next prove that any automorphism preserving transpositions is an inner automorphism. Regard a transposition as an edge of the complete graph on the $n$ letters. Two distinct transpositions fail to commute exactly when their edges meet. Any family of pairwise intersecting edges is contained in a vertex star or a triangle: after choosing $\{a,b\}$ and $\{a,c\}$, an edge not containing $a$ must be $\{b,c\}$, which precludes any fourth edge. For $n\ge 5$, the stars are therefore distinguished as the maximal such families of size $n-1\ge 4$. An automorphism permutes stars, inducing a permutation $\pi$ of the letters; their pairwise intersections show that $(ij)$ is sent to $(\pi (i)\pi (j))$. Since transpositions generate $S_n$, this is conjugation by a permutation.

For $n=4$, stars and triangles both have three edges, but a star generates $S_4$ while a triangle generates a subgroup $S_3$. An automorphism preserves the order of the generated subgroup and hence still distinguishes stars. For $n=3$, every permutation of the three transpositions is induced by conjugation in $S_3$, since its conjugation action on them is faithful and has order six. The groups $S_1,S_2$ have trivial automorphism groups. This completes the small cases.

It follows that all automorphisms for $n\ne 6$ are inner. In degree six, an automorphism permutes the two involution classes of size 15; the kernel of this action consists precisely of the inner automorphisms. Therefore the automorphisms of the symmetric group satisfy

For a profinite group, its profinite Frattini subgroup isUse $G$ for an empty intersection. This is a closed normal subgroup, invariant under continuous group automorphisms, since these automorphisms permute the maximal open subgroups. Let $H=\overline{⟨X⟩}$. If $H=G$, its image certainly topologically generates $G/\Phi (G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the open normal subgroup basis of a profinite group gives $N{⊴}_{o}G$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed subgroup $H$. The proper subgroup $HN/N$ of the finite group $G/N$ lies in a maximal proper subgroup. Its preimage $M$ is a maximal proper open subgroup of $G$, contains $H$, and contains $\Phi (G)$. Since $H\Phi (G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi (G)$ would force $H\Phi (G)=G$, contradicting $H\Phi (G)\le M<G$. Generation can therefore be tested modulo the profinite Frattini subgroup.

For a finite p-group, every maximal proper subgroup is normal and has index $p$. One proof uses the normalizer condition for finite p-groups: maximality forces its normalizer to be the whole group, and its simple $p$-group quotient has order $p$. Therefore every such maximal subgroup contains powers and group commutators. Conversely, the quotient by the algebraic $G^p[G,G]$ is an elementary abelian p-group; its nonzero elements are separated by linear maps to ${\mathbb{F}}_p$. Their kernels give maximal subgroups. This proves the finite Frattini subgroup formula.

Now take a pro-p group and put $D=\overline{G^p[G,G]}$, retaining the closure of the algebraically generated subgroups. A maximal open subgroup is the preimage of a maximal subgroup of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the profinite group $G/D$ detects $gD$. That quotient is elementary abelian, and a linear functional nonzero at $gD$ gives a continuous group homomorphism $G\to {\mathbb{F}}_p$ whose kernel is a maximal open subgroup missing $g$. HenceThe closure bar is present in the original PDF; the converted TeX fraction is an extraction error.

For the finite-index assertion, first pass from the arbitrary abstract subgroup $K$ to its normal core of a subgroup $C$. The action on the finitely many cosets of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^{r}q$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By Lagrange theorem, $X_m\subseteq C$. For $N{⊴}_{o}G$, the quotient $G/N$ is a finite p-group of order, say, $p^s$. Choose an integer $a$ with $aq\equiv 1(\mathrm{mod}{p}^s)$. Thenso $g^{p^r}\in X_{m}N$ for every such $N$. The set $X_m$ is the continuous image of the compact space $G$, hence is closed. Consequently $g^{p^r}\in {\bigcap }_N{X}_{m}N=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By Cauchy theorem for groups, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. This proof does not assume continuity of the finite quotient map.

To prove openness under finite generation, first establish commutator collection with fixed generators in nilpotent groups. If a nilpotent group $P$ is generated by $a_1,\dots ,a_d$, every element of $P\prime$ isInduct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/{\gamma }_c(P)$; the remaining error lies in the central subgroup ${\gamma }_c(P)=[{\gamma }_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as ${\prod }_i[y_i,a_i]$ with $y_i\in {\gamma }_{c-1}(P)$. Also $[P\prime ,{\gamma }_{c-1}(P)]\le {\gamma }_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_i{y}_i$ multiplies the desired product by precisely that central error, proving the formula.

Let $a_1,\dots ,a_d$ now topologically generate $G$. Every finite continuous quotient is a nilpotent group, so the formula holds there with these fixed generators. For $z\in \overline{[G,G]}$, the sets of tuples $(x_1,\dots ,x_d)\in G^d$ solving $z\equiv {\prod }_i[x_i,a_i](\mathrm{mod}N)$ are nonempty closed subsets of a compact space. They have the finite intersection property, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic commutator subgroup is the compact image of $G^d$ under this product map, and is closed.

Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian pro-p group. The continuous map ${\mathbb{Z}}_p^d\to A$ taking $({\lambda }_i)$ to ${\prod }_i{\bar{a}}_i^{{\lambda }_i}$ is onto, because its compact image contains a dense generating subgroup. Therefore $A/pA$ is generated as an elementary abelian group by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed cosets. The finite-index result and the maximal-subgroup property of finite p-groups show that any abstract index-$p$ subgroup of $G$ is normal. It contains this open subgroup, so is itself open.

Finally, a finite-index normal subgroup $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ subgroup in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open subgroup of a finitely generated pro-p group is finitely generated: intersect a finitely generated dense abstract subgroup with $H$, use the Schreier lemma, and observe that the intersection is dense in the open subgroup $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the normal core of a subgroup, $K$ is a union of open cosets of $C$. Every finite-index subgroup of a finitely generated pro-p group is open.

A topological group module $A$ is a topological abelian group with a jointly continuous action of $G$ by group automorphisms. Its continuous cochains with topological coefficients arewith pointwise addition. The inhomogeneous differential isFor $n=0$, this means $(da)(g)=ga-a$. Continuity of multiplication and of the action ensures that $d$ preserves continuous cochains. Expanding $d(dc)$ gives cancellation in pairs: the two orders of multiplying neighbouring arguments agree by associativity, and the two ways of applying the action agree by its action law. Thus $d^2=0$. The cocycles are $Z^n=\ker (d:C^n\to C^{n+1})$ and the coboundaries are $B^n=\mathrm{im}(d:C^{n-1}\to C^n)$, with $B^0=0$. DefineFor general topological coefficients this is an algebraic quotient of the continuous cochains; its optional quotient topology need not be Hausdorff. For discrete coefficients it is the usual continuous cohomology of a profinite group. In degree zero it is the subgroup of invariant elements.

Give the discrete prime field ${\mathbb{F}}_p$ its trivial $G$-action. The one-cocycle condition becomes $c(gh)=c(g)+c(h)$ and every one-coboundary is zero. HenceEvery such group homomorphism kills powers and group commutators, hence their closure $\Phi (G)$, and every continuous functional on the quotient pulls back. If $G$ is topologically finitely generated, $V=G/\Phi (G)$ is a finite elementary abelian group. To verify finiteness without assuming it, the closure of a subgroup generated by $r$ elements of this abelian exponent-$p$ quotient has at most $p^r$ elements; a finite subgroup is already closed. Thus $V$ is a finite-dimensional vector space. Question 2 says that a generating family of $G$ is minimal precisely when its images give a basis of $V$. The dual space has the same finite dimension. This proves

There is a genuine missing hypothesis if the printed assertion is read for arbitrary pro-p groups and $d(G)$ means the least cardinality of any dense generating family. Here is an explicit infinite-rank failure of the first-cohomology generator formula. Let $I=\{0,1{\}}^{\mathbb{N}}$ and $G={\prod }_{r\in I}{\mathbb{F}}_p$. Enumerate the finite binary strings as $s_1,s_2,\dots$ and define $g_n(r)=1$ if $s_n$ is a prefix of $r$, and $0$ otherwise. For finitely many distinct sequences $r_1,\dots ,r_k$, choose long enough prefixes to distinguish each individually. The corresponding $g_n$ project onto the coordinate unit vectors of ${\mathbb{F}}_p^k$, so $⟨g_n:n\ge 1⟩$ surjects onto every finite coordinate projection and is dense in the product topology. No finite family can generate densely, since a finitely generated abelian exponent-$p$ subgroup is finite and closed. Thus $d(G)={\aleph }_0$. On the other hand, continuity of a linear functional to ${\mathbb{F}}_p$ means its kernel contains a basic neighbourhood, so it depends on finitely many coordinates. The coordinate projections are consequently a basis of the continuous dual space, givingThe unrestricted infinite-rank formula is false. A convention requiring a generating basis tending to the identity is different from the printed unrestricted topological generator count; the finite-rank identity proved above has no such ambiguity.

For finite $X$, a free pro-p group $F_p(X)$ is characterized by the universal property that every map $X\to P$ to a pro-p group extends uniquely to a continuous group homomorphism $F_p(X)\to P$. It is the pro-p completion of the abstract free group on $X$: each map to a finite $p$-group factors through one of the defining finite quotients, and passing to the inverse limit gives the extension to any pro-p group; uniqueness follows from density. This unrestricted universal property can also define the completion on an arbitrary set $X$. When using an infinite basis required to tend to the identity, one instead imposes the universal property only for maps tending to the identity. Only the finite-rank version is needed for the following argument.

A finite pro-p presentation is an isomorphismThere are finitely many generators and finitely many relators, and the denominator is their closed normal closure. The relators are elements of the free pro-p group; they may be limits of ordinary words. That distinction is important when writing conjugation relations below.

Now suppose $K{⊴}_{o}G$ has a finite pro-p presentation with generators $k_1,\dots ,k_a$ and relators $R_1,\dots ,R_t$. The quotient $Q=G/K$ is a finite p-group. Choose a finite finite pro-p presentation $Q=⟨{\bar{y}}_1,\dots ,{\bar{y}}_b\mid S_1,\dots ,S_c⟩$. Such a presentation exists, for example by using its entire finite multiplication table. Lift the quotient generators to $y_j\in G$. For every $i,j$, choose elements $W_{ij},V_{ij}$ of the free pro-p group on the $x_i$ that map respectively to $y_j^{-1}{k}_i{y}_j$ and $y_j{k}_i{y}_j^{-1}$ in $K$. Choose $U_{\ell }(x)$ mapping to $S_{\ell }(y)$, which lies in $K$. Form the finite finite pro-p presentationAll listed relations hold in $G$, giving a continuous surjective group homomorphism $P\to G$: its compact image contains $K$ and maps onto $Q$.

Let $X$ be the closed subgroup of $P$ topologically generated by the $x_i$. Both conjugation directions were included, so each $y_j$ normalizes $X$. The $x_i$ also normalize it, and its closed normalizer contains the generators of $P$. Thus $X⊴P$. The relations $R_h$ give a continuous surjection $K\to X$. Its composite with $X\to K$ induced by $P\to G$ fixes every $k_i$, and hence is the identity. Therefore the surjection $K\to X$ is injective as well, so $X\cong K$. On quotienting by $X$, the conjugation and kernel relations disappear, and the remaining presentation is exactly $⟨y_1,\dots ,y_b\mid S_1,\dots ,S_c{⟩}_{\mathrm{pro}\text{-}p}=Q$. Thus $P/X\to G/K$ is an isomorphism. The kernel of $P\to G$ lies in $X$ and meets $X$ trivially, so is trivial. Consequently $P\cong G$, proving

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.

Put $n=|\Omega |\ge 2$, and let $H=G_{\alpha }$. Sharp two-transitivity gives $|G|=n(n-1)$ and $|H|=n-1$. A nonidentity element fixes at most one point. Counting the nonidentity elements in the $n$ stabilizer subgroups shows that there are $n-1$ fixed-point-free elements. Let $N$ be this set together with the identity. We first prove it is a normal subgroup, rather than presuming that fixed-point-free elements are closed under multiplication.

For $n=2$, $G=C_2$ and $N=G$ directly. Otherwise use complex characters of a finite group. Let $\pi$ be the permutation character and $\psi =\pi -1$, the character of the permutation representation with its constant line removed. For every nontrivial irreducible character $\theta$ of $H$, form the virtual characterIts values are $\theta (1)$ at the identity and at every fixed-point-free element. At an element with one fixed point, conjugate it to $h\in H\setminus \{1\}$; induction gives ${\phi }_{\theta }(g)=\theta (h)$, because there is exactly one fixed coset.

The identity and the $n-1$ fixed-point-free elements together contribute $n\theta (1)\overline{\eta (1)}$ to the inner product. The remaining elements are partitioned into the nonidentity parts of the $n$ stabilizer subgroups. Hence character orthogonality givesA virtual character of norm one is plus or minus an irreducible character: its coefficients in the irreducible-character basis are integers whose squares sum to one. Its positive degree $\theta (1)$ selects the plus sign. Thus each ${\phi }_{\theta }$ is an actual irreducible character.

For a group representation, $\chi (g)=\chi (1)$ holds exactly on its kernel: make the representation unitary and compare the sum of its unit-modulus eigenvalues with its dimension. All of $N$ therefore lies in the intersection of the kernels of the ${\phi }_{\theta }$. Conversely, a nonidentity element fixing a point gives $h\ne 1$ in $H$. Some nontrivial irreducible character of $H$ has $\theta (h)\ne \theta (1)$, since otherwise the regular representation of $H$ would not vanish at $h$. ConsequentlyThis proves normality and subgroup closure. It has order $n$ and no nonidentity element fixing a point, so it is a regular permutation subgroup.

Now $H$ acts transitively by conjugation on $N\setminus \{1\}$: identify an element of $N$ with its image of $\alpha$ and use transitivity of $H$ on the remaining points. Thus all nonidentity elements of $N$ have the same order. Taking a suitable power of one element shows this common order is a prime $p$. By Cauchy's theorem no other prime divides $|N|$, so $N$ is a $p$-group. Its nontrivial center is $H$-invariant, so transitivity forces the center to be all of $N$. Therefore $N$ is elementary abelian of order $p^d=n$.

Since $|H|=p^d-1$ is prime to $p$, $N$ is the unique Sylow $p$-subgroup of $G$. Uniqueness makes it characteristic under every group automorphism. We have provedThe character argument supplies the regular kernel of a finite sharply two-transitive group; the final Sylow argument establishes the stronger characteristic assertion.

Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

For fixed $g\in G$, conjugation $g_{\ast }:x\mapsto gx{g}^{-1}$ preserves multiplication:and its inverse is $(g^{-1}{)}_{\ast }$. It is therefore a group automorphism. Furthermore,so $\theta :g\mapsto g_{\ast }$ is a homomorphism $G\to \mathrm{Aut}(G)$. Its kernel consists of elements commuting with every $x$, hence

For $\alpha \in \mathrm{Aut}(G)$,so the inner automorphism group is normal in the automorphism group.

If $G=⟨x⟩$ is cyclic, every automorphism has the form $x\mapsto x^a$. Two such maps commute because $ab=ba$, so $\mathrm{Aut}(G)$ is abelian. The conclusion fails for general abelian groups: for $G=C_2\times C_2$, its three nonidentity elements may be permuted arbitrarily, givingwhich is nonabelian.

For $h\in H$, define $\pi (h)(n)=hn{h}^{-1}$. Since $N$ is a normal subgroup, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an group automorphism of $N$. MoreoverThus $\pi :H\to \mathrm{Aut}(N)$ is a group homomorphism, induced by the group action of conjugation.

If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Henceis a group homomorphism, since $(n_1{h}_1)(n_2{h}_2)=n_1{n}_2{h}_1{h}_2$. Its image of a group homomorphism is a subgroup containing both $N$ and $H$, and every product $nh$ lies in $⟨N,H⟩$, so its image is exactly $⟨N,H⟩$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives

Write $C_p=⟨g⟩$. A group homomorphism from this cyclic group is determined by the image of $g$, because $\varphi (g^j)=\varphi (g{)}^j$. Each choice $\varphi (g)=g^a$, with $a=0,\dots ,p-1$, gives a well-defined group homomorphism. It is an group automorphism exactly when $g^a$ generates $C_p$.

For a prime number $p$, every $a=1,\dots ,p-1$ is coprime to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial group homomorphism and fails to be bijective. Consequently

A map $\varphi :G\to G$ is a group isomorphism if it is a bijection and a group homomorphism, meaningSuch a map from a group to itself is an group automorphism. Its inverse is also a group homomorphism: if $u=\varphi (x)$ and $v=\varphi (y)$, then ${\varphi }^{-1}(uv)=xy={\varphi }^{-1}(u){\varphi }^{-1}(v)$.

Assume first $n\ge 2$. All transpositions form a single conjugacy class $T_1$. A group automorphism preserves order of a group element, so the image of a transposition is a nonidentity involution. By the order formula, its disjoint cycle decomposition contains only one-cycles and two-cycles, with at least one two-cycle. Thus it lies in some $T_k$, where $1\le k\le \lfloor n/2\rfloor$.

Moreover an automorphism maps the entire conjugacy class onto the conjugacy class of the image: for a fixed transposition $t$,and $\varphi (g)$ runs over all of $S_n$ by surjectivity. Part (a)(i) identifies that latter conjugacy class as $T_k$. Hence $\boxed{\varphi (T_1)=T_k}$, with equality rather than just inclusion. If $n=1$, $T_1$ is empty and the equality holds trivially with $k=1$.

A group homomorphism preserves multiplication, $\varphi (ab)=\varphi (a)\varphi (b)$. A group isomorphism is a bijective homomorphism. An isomorphism from a group to itself is a group automorphism.

The automorphism group contains the identity map. Composition is associative, and if $\varphi ,\psi$ are automorphisms, $(\varphi \circ \psi )(ab)=\varphi (\psi (a)\psi (b))=(\varphi \circ \psi )(a)(\varphi \circ \psi )(b)$; composition is also bijective. For $u=\varphi (a)$ and $v=\varphi (b)$, ${\varphi }^{-1}(uv)={\varphi }^{-1}(\varphi (ab))=ab={\varphi }^{-1}(u){\varphi }^{-1}(v)$, proving the inverse bijection is a homomorphism. These facts establish all group axioms.

A finite cyclic group is $G=⟨g⟩=\{e,g,\dots ,g^{n-1}\}$ with $g$ of order $n$. The map $g^k\mapsto e^{2\pi ik/n}$ is well defined modulo $n$, bijective onto the roots of unity, and preserves multiplication. It identifies $G$ with the multiplicative group of $n$th roots of unity.

For prime $n$, Lagrange's theorem leaves only subgroup orders one and $n$: there is no nontrivial proper subgroup. The trivial subgroup $\{e\}$ is nevertheless proper when $n>1$, so that qualification is necessary in the printed claim. The characterization of groups with no proper nontrivial subgroups makes the distinction explicit.

For composite $n$, choose a prime divisor $p$ and take $\boxed{H_0=\{e\},H_1=⟨g^p⟩}$, of orders $1$ and $n/p$. They are distinct proper subgroups, with $H_1$ nontrivial. If the question intended two nontrivial proper subgroups, that demand is not always possible: a cyclic group of order $p^2$ has only the one of order $p$. The subgroups and quotients of a cyclic group are controlled by its divisors. The order-one group is a separate trivial case, with no proper subgroups.

The two products are conjugate:Conjugation by $a^{-1}$ is a group automorphism, so part (a) provesDirectly, cancellation gives $(ba{)}^r=a^{-1}(ab{)}^{r}a$ for every positive integer $r$. Thus a power of one product is the identity exactly when the same power of the other is the identity. This proves that conjugation preserves element order, even without the finiteness assumption.

Use the fact that $K$ is a normal subgroup to move a factor of $K$ past a factor of $H$. For $h_1,h_2\in H$ and $k_1,k_2\in K$,whileThe identity element belongs to $HK$. Hence $HK$ is a subgroup, by the subgroup test. This proves the product of a subgroup with a normal subgroup construction; $H$ need not be a normal subgroup.

For each $h\in H$, conjugation maps $K$ to itself and respects its multiplication. The inverse of $k\mapsto h^{-1}kh$ is $k\mapsto hk{h}^{-1}$, so both are group automorphisms of $K$.

The printed formula requires a composition convention. For the printed ${\varphi }_h(k)=h^{-1}kh$, ordinary function composition givesIndeed, $(h_1{h}_2{)}^{-1}k(h_1{h}_2)=h_2^{-1}(h_1^{-1}k{h}_1)h_2$. Thus the displayed assignment is a group antihomomorphism with the usual multiplication in the automorphism group. It is a group homomorphism if the automorphisms are written as right operators and multiplied in the corresponding reverse order; this is precisely why right conjugation reverses composition.

With the usual left function composition, the corresponding group homomorphism is insteadDirect expansion gives ${\psi }_{h_1{h}_2}(k)=h_1(h_{2}k{h}_2^{-1})h_1^{-1}=({\psi }_{h_1}\circ {\psi }_{h_2})(k)$. This states the intended conjugation action without silently reversing the order in the printed formula.

For a normal subgroup $K$ of $G$, each $c_h:k\mapsto h^{-1}kh$ is a group automorphism of $K$. Its inverse is $c_{h^{-1}}$, but the dependence on $h$ is a group antihomomorphism under ordinary function composition, because $(gh{)}^{-1}k(gh)=h^{-1}(g^{-1}kg)h$. The assignment $h\mapsto (k\mapsto hk{h}^{-1})$ instead is a group homomorphism to the automorphism group. Equivalently the first convention is a right group action, while the second is a left group action.

A permutation of order two is a product of $k$ disjoint transpositions. Its conjugacy class has the displayed size. For $k\ge 2$, equality with the transposition class requires $(n-2)!/(n-2k)!=2^{k-1}k!$. The left side grows strictly with $n$. At $n=2k$, its ratio to the right side is one only for $k=3$; at $k=2$ it jumps from below to above one between $n=4$ and $n=5$, and for $k\ge 4$ it is already greater than one. Thus degree six is the only possible exceptional degree for an group automorphism not preserving transpositions.